BIOS 625 HW 5 Solutions Sheet - dbandyop/BIOS625/HW/hw52015_Soln.pdf · BIOS 625 HW #5 Solutions...

BIOS 625 HW #5 Solutions Sheet

November 10, 2015

Problem 1. Agresti 5.19.

R = 1 : logit(π̂) = 6.7 + .1A+ 1.4S. R = 0 : logit(π̂) = 7.0 + .1A+ 1.2SThe YS conditional odds ratio is exp(1.4) = 4.1 for blacks and exp(1.2) = 3.3for whites. Note that .2, the coeff. of the cross-product term, is the differencebetween the log odds ratios 1.4 and 1.2. The coeff. of S of 1.2 is the logodds ratio between Y and S when R = 0 (whites), in which case the RSinteraction does not enter the equation. The P-value of P < .01 for smokingrepresents the result of the test that the log odds ratio between Y and S forwhites is 0.

Problem 2. Agresti 5.20

Part a. The estimated log odds ratio between race and driving after con-suming a substantial amount of alcohol was −.72 in Grade 12 (i.e., for eachgender, the estimated odds for blacks of driving after consuming a substan-tial amount of alcohol were e−.72 = .49 times the estimated odds for whites.The corresponding estimated log odds ratio was −.72 + .74 = .02 for Grade9, −.72 + .38 = .34 for Grade 10, and −.72 + .01 = −.71 for Grade 11. i.e.there is essentially no association in Grade 9, but the association changes toan odds ratio of about .5 in Grades 11 and 12.


Are people with more social ties less likely to get colds? Use logistic modelsto analyze the 2x2x2x2 contengency table on pp. 1943 of article by S. Cohenet al., J.Am.Med.Assoc. 277 (24).

See next several pages of SAS output:

HW5 Problem 3

Residuals vs predicted eta_i with LOESS Overlay

The LOESS Procedure

Selected Smoothing Parameter: 1

Dependent Variable: res

04:23 Tuesday, November 10, 2015 1

Stepwise Model Selection

Response Profile

Ordered

Value

Binary

Outcome

Total

Frequency

1 Event 109

2 Nonevent 167

Stepwise Selection Procedure

Class Level Information

Class Value

Design

Variables

titer f<=2f 1

f>=4f 0

virus fHanks 1

fRV39f 0

social f1-5f 1

f>=6f 0

Analysis of Maximum Likelihood Estimates

Parameter DF Estimate

Standard

Error

Wald

Chi-Square Pr > ChiSq

Intercept 1 -1.5530 0.2880 29.0752 <.0001

titer f<=2f 1 1.9280 0.2923 43.5231 <.0001

virus fHanks 1 -0.6051 0.2805 4.6533 0.0310

social f1-5f 1 0.6538 0.2782 5.5228 0.0188

Odds Ratio Estimates

Effect

Point

Estimate

95% Wald

Confidence Limits

titer f<=2f vs f>=4f 6.876 3.878 12.193

virus fHanks vs fRV39f 0.546 0.315 0.946

social f1-5f vs f>=6f 1.923 1.115 3.317

HW5 Problem 3


The LOESS Procedure




Hosmer and Lemeshow

Goodness-of-Fit Test

Chi-Square DF Pr > ChiSq

2.1753 6 0.9029

-2 -1 0 1

Value of the Linear Predictor

-1.0

-0.5

0.0

0.5

1.0

res

HW5 Problem 3

Residuals vs predicted eta_i

-2 -1 0 1


-1.0

-0.5

0.0

0.5

1.0

res

Fit Plot for res

-2 -1 0 1


-1.0

-0.5

0.0

0.5

1.0

res

Smooth = 1

Fit Plot for res

HW5 Problem 3


The LOESS Procedure




-2 -1 0 1


-1.0

-0.5

0.0

0.5

1.0

Resi

dual

Residual Plot for res

Fit Diagnostics for res

LinearInterpolation

8Fit Points

4.6142Residual SS

1Degree

8Local Points

1Smooth

8Observations

-1.0 -0.5 0.0 0.5 1.0

Predicted Value

-1.0

-0.5

0.0

0.5

1.0

res

Proportion Less

0.2 0.8

Residual

0.2 0.8

Fit–Mean

-1.0

-0.5

0.0

0.5

1.0

-1.5 -0.5 0.5 1.5

Quantile

-1.0

-0.5

0.0

0.5

1.0

Resid

ual

-2.7 -1.5 -0.3 0.9 2.1

Residual

0

10

20

30

40

50

Perc

en

t

-0.2 -0.1 0.0 0.1

Predicted Value

-1.0

-0.5

0.0

0.5

1.0

Resid

ual


Std. Pearson residual plots

HW5 Problem 3

-2 -1 0 1


-1.0

-0.5

0.0

0.5

1.0

Sta

ndard

ized P

ears

on R

esi

dual

-2 -1 0 1


-1.0

-0.5

0.0

0.5

Pears

on R

esi

dual


The derivative equals βeα+βx

[1+eα+βx]2= βπ(x)(1− π(x))


The odds ratio eβ is approximately equal to the relative risk when the prob-ability is near 0 and the complement is near 1, since

eβ = [π(x+ 1)/(1− (π(x+ 1)]/[π(x)/(1− π(x+ 1))] ≈ π(x+ 1)/π(x)

Problem 6. Finney and Pregibon vasoconstriction

a. The AICs for the binary regressions using logit, probit and complimen-tary log-log links are 35.227, 35.287 and 32.622, respectively. Therefore, thecomplimentary log-log model has the smallest AIC.

b. In the fitted complimentary log-log model, we have β̂0 = −2.9736,β̂1 = 3.9702, and β̂2 == 4.3361. Therefore:

P (V = 1) = 1−exp[−exp{−2.9736+3.9702log(volume)+4.3361log(rate)}]P (V = 1) = 1− exp[−0.051 ∗ volume3.9702 ∗ rate4.3361]

Thus, increasing volume or rate increases the probability of vaso constric-tion.

c. The Hosmer-Lemeshow test statistic is 10.0705 (with d.f. = 8). Thiscorresponds to a p-value of 0.2601. Therefore, there is no evidence of lack-of-fit.

d. An approach to detect ill-fit observations is to consider observationswhere the absolute value of the Pearson Residual is greater than 3. Usingthis approach, there are 2 observations that are considered ill-fit.

e. There are two influential observations according to the Dfbetas andCis, and they are the same observations that have large Pearson residuals.

f. When we remove observations 4 and 18, the AIC value for the modeldrops from 32.622 to 13.32. None of the standardized Pearsons residu-als are greater than 2.5, suggesting that no observations are especially ill-fitting. However, the coefficients change significantly. In particular, beforewe had β̂0 = −2.9736, β̂1 = 3.9702, and β̂2 == 4.3361. Now we haveβ̂0 = −16.58384, β̂1 = 21.02387, and β̂2 = 25.30425.



HW5, problem 6

Vaso Data Analysis [Cloglog]

Probability modeled is cons=1.

Model Fit Statistics

Criterion Intercept

Only

Intercept

and

Covariates

AIC 56.040 32.622

SC 57.703 37.613

-2 Log L 54.040 26.622



Standard

Error

Wald


Intercept 1 -2.9736 1.0934 7.3960 0.0065

lvol 1 3.9701 1.3825 8.2469 0.0041

lrate 1 4.3361 1.5589 7.7369 0.0054

Hosmer and Lemeshow



10.0705 8 0.2601


HW5, problem 6

Vaso Data Analysis [Cloglog]

The LOGISTIC Procedure

Regression Diagnostics

Case

Number

Covariates

Pearson

Residual

Deviance

Residual

lvol

DfBeta

lrate

DfBeta

Confidence

Interval

Displacement

C

Confidence

Interval

Displacement

CBar lvol lrate

4 0.4055 -0.2877 3.6225 2.3012 -0.8722 -0.9173 1.0287 0.9587

18 0.3471 -0.1625 3.0796 2.1679 -0.7663 -0.7754 0.7999 0.7419

Influence Diagnostics

01cons

0.0

0.2

0.4

0.6

0.8

1.0

CI

Dis

pla

cem

en

ts C

0.0

0.1

0.2

0.3

Lev

era

ge

-1

0

1

2

Dev

ian

ce R

esid

ual

-1

0

1

2

3

Pears

on

Resid

ual

0 10 20 30 40

Case Number

0 10 20 30 40

Case Number



01cons

0

2

4

6

Dev

ian

ce D

ele

tio

n D

iffe

ren

ce

0.0

2.5

5.0

7.5

10.0

12.5

Ch

i-sq

uare

Dele

tio

n D

iffe

ren

ce

0.0

0.2

0.4

0.6

0.8

1.0

CI

Dis

pla

cem

en

ts C

Bar

0 10 20 30 40

Case Number

0 10 20 30 40

Case Number


01cons

DfB

eta

s

lratelvol

Intercept

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

0 10 20 30 40

Case Number

0 10 20 30 40

Case Number


HW5, problem 6

Vaso Data Analysis [Cloglog] Without Observations 4 and 18

Number of Observations Read 37

Number of Observations Used 37

Response Profile

Ordered

Value cons

Total

Frequency

1 1 18

2 0 19

Probability modeled is cons=1.

Model Fit Statistics

Criterion

Intercept

Only

Intercept

and

Covariates

AIC 53.266 13.325

SC 54.877 18.158

-2 Log L 51.266 7.325

Testing Global Null Hypothesis: BETA=0

Test Chi-Square DF Pr > ChiSq

Likelihood Ratio 43.9410 2 <.0001

Score 18.7712 2 <.0001

Wald 3.1492 2 0.2071



Standard

Error

Wald


Intercept 1 -16.5838 10.3652 2.5598 0.1096

lvol 1 21.0237 13.0935 2.5782 0.1083

lrate 1 25.3041 16.7511 2.2819 0.1309

Hosmer and Lemeshow



0.6550 5 0.9853


For table 10.1, treating marijuana; Set the value as 1 if the predictor use ofalcohol is YES and 0 otherwise; 1 if the predictor use of cigarette is YES and0 otherwise; female as 1 and male as 0; white as 1 and 0 other race. Using abackwards elimination, the final model is composed of the predictors alcohol,cigarette and gender. All interaction terms were non-significant. Therefore,the fitted model is:

logit(π̂) = −5.1883+ 3.0201 ∗ alcohol+2.8591 ∗ cigarette− 0.3279 ∗ gender

The Pearson GOF statistic yields a p-value of 0.8781, which indicates theresno evidence of gross lack of fit in this model. The odds of marijuana useamong alcohol users is exp(3.0201) = 20.494 times of the odds among non-alcohol users when keeping the remaining parameters constant; the odds ofmarijuana use among smokers is exp(2.8591) = 17.446 times of the oddsamong non-smokers when keeping the remaining parameters constant. Andthe odds of marijuana use among males is 1/exp(0.3279) = 1.388 times theodds among females when keeping the remaining parameters constant.



Problem 7, Agresti 6.4

Stepwise Regression on Marijuana Data

Final Model Selected


Parameter DF Estimate Standard

Error

Wald


Intercept 1 -5.1883 0.4769 118.3642 <.0001

a 1 1 3.0201 0.4653 42.1249 <.0001

c 1 1 2.8591 0.1642 303.0914 <.0001

g 1 1 -0.3279 0.1026 10.2200 0.0014


Effect

Point

Estimate

95% Wald

Confidence Limits

a 1 vs 2 20.494 8.233 51.016

c 1 vs 2 17.446 12.645 24.071

g 1 vs 2 0.720 0.589 0.881

Hosmer and Lemeshow



1.8966 3 0.5941


-6 -4 -2 0


-1

0

1

2

res


Residuals vs predicted eta_i

-6 -4 -2 0


-1

0

1

2

res

Fit Plot for res

-6 -4 -2 0


-1

0

1

2

res

Smooth = 0.969

Fit Plot for res


-6 -4 -2 0


-1

0

1

2

Resi

dual

Residual Plot for res

Fit Diagnostics for res

LinearInterpolation

8Fit Points

11.903Residual SS

1Degree

15Local Points

0.9688Smooth

16Observations

-1 0 1 2

Predicted Value

-1

0

1

2

res

Proportion Less

0.0 0.4 0.8

Residual

0.0 0.4 0.8

Fit–Mean

-1

0

1

2

-2 -1 0 1 2

Quantile

-2

-1

0

1

2

Resid

ual

-2.8 -1.2 0.4 2

Residual

0

10

20

30

40

50

Perc

en

t

-0.1 0.0 0.1

Predicted Value

-1

0

1

2

Resid

ual


Std. Pearson residual plots


-6 -4 -2 0


-1

0

1

2

Sta

ndard

ized P

ears

on R

esi

dual

-6 -4 -2 0


-1.0

-0.5

0.0

0.5

1.0

Pears

on R

esi

dual

Problem 8. Dixon and Massey

First, we fit the model with all main effects and 2-way interactions. Choosingthe predictors that have p-value greater than 0.05, we get a model with thepredictors age and weight. The fitted model is given by:

logit(π̂) = β̂0 + β̂1 ∗Age+ β̂2 ∗ weight

logit(π̂) = −7.5128 + 0.0636 ∗Age+ 0.0160 ∗ weightA backward elimination can also be done to find the best model. TheHosmer-Lemeshow test has a p-value equal to 0.7941, which indicates thatthere is no gross lack of fit in this model. The odds of having an inci-dent increases by a multiplicative factor of exp(0.0636) = 1.066 (95%C.I is(1.025,1.108))for every unit increase in age while holding weight constant;and the odds the odds of having an incident increases by a multiplicativefactor of 1.016 95%C.I is (1.000,1.032) for every unit increase in weight whileholding age constant.



Problem 8, heart.sas

Stepwise Regression on Heart Data


Parameter DF Estimate Standard

Error

Wald


Intercept 1 -7.5128 1.7093 19.3176 <.0001

AG 1 0.0636 0.0197 10.4389 0.0012

W 1 0.0160 0.00795 4.0464 0.0443


Effect

Point

Estimate

95% Wald

Confidence Limits

AG 1.066 1.025 1.108

W 1.016 1.000 1.032

Hosmer and Lemeshow



4.6510 8 0.7941


Backward Regression on Heart Data



Standard

Error

Wald


Intercept 1 -7.5128 1.7093 19.3176 <.0001

AG 1 0.0636 0.0197 10.4389 0.0012

W 1 0.0160 0.00795 4.0464 0.0443


Effect Point

Estimate

95% Wald

Confidence Limits

AG 1.066 1.025 1.108

W 1.016 1.000 1.032

Hosmer and Lemeshow



4.6510 8 0.7941


Cook's distance and Std. Pearson resids


0 50 100 150 200

ID

0.00

0.05

0.10

0.15

0.20

Confi

dence I

nte

rval D

ispla

cem

ent

C

0 50 100 150 200

ID

0

2

4

Sta

ndard

ized P

ears

on R

esi

dual

-4 -3 -2 -1 0


0

2

4

Sta

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ized P

ears

on R

esi

dual



Cook's distance and Std. Pearson resids

Obs ID AG S D Ch H W eta phat lcl ucl p h2 q c

21 21 37 110 70 312 71 170 -2.43996 0.080176 0.044684 0.13973 3.38713 0.007438 3.39979 0.08662

29 29 40 110 74 336 68 166 -2.31323 0.090033 0.053777 0.14694 3.17915 0.006555 3.18962 0.06713

41 41 40 130 90 520 68 169 -2.26522 0.094045 0.056668 0.15210 3.10375 0.006636 3.11410 0.06478

86 86 34 110 80 214 67 139 -3.12677 0.042017 0.017700 0.09646 4.77495 0.008292 4.79487 0.19223

126 126 28 120 86 386 70 189 -2.70816 0.062494 0.025570 0.14481 3.87319 0.013255 3.89912 0.20423

159 159 40 110 70 244 70 161 -2.39324 0.083690 0.048877 0.13966 3.30891 0.006602 3.31989 0.07325

184 184 43 138 94 320 65 159 -2.23450 0.096695 0.059371 0.15365 3.05643 0.006345 3.06618 0.06003

55

19

5190

123

81

44 54

129

124

188 42

151

177 184

41

29

159

21

126

86

0.00

0.25

0.50

0.75

1.00

Sensi

tivity

0.00 0.25 0.50 0.75 1.00

1 - Specificity

Points labeled by observation number

ROC Curve for ModelArea Under the Curve = 0.7349


0.00

0.25

0.50

0.75

1.00

Pro

babili

ty

20 40 60 80

AG

PredictedObserved

Predicted Probabilities for CNT=1 with 95% Confidence LimitsAt W=165.2

R-Square 0.0759 Max-rescaled R-Square 0.1410


0.00

0.25

0.50

0.75

1.00

Sensi

tivity

0.00 0.25 0.50 0.75 1.00

1 - Specificity

ROC Curve for ModelArea Under the Curve = 0.7349

ROC Model: Age



Standard

Error

Wald


Intercept 1 -4.7925 0.9595 24.9481 <.0001

AG 1 0.0633 0.0192 10.8270 0.0010


Effect Point

Estimate

95% Wald

Confidence Limits

AG 1.065 1.026 1.106


0.40.34

0.32

0.31

0.28

0.25

0.22

0.21 0.18

0.17

0.160.15

0.14

0.13

0.12

0.11

0.11

0.09

0.08

0.07

0.05

0.00

0.25

0.50

0.75

1.00

Sensi

tivity

0.00 0.25 0.50 0.75 1.00

1 - Specificity

Points labeled by predicted probability

ROC Curve for AgeArea Under the Curve = 0.6985

ROC Model: Weight



Standard

Error

Wald


Intercept 1 -4.7244 1.3692 11.9064 0.0006

W 1 0.0167 0.00783 4.5536 0.0328


Effect Point

Estimate

95% Wald

Confidence Limits

W 1.017 1.001 1.033


0.35

0.25

0.23

0.19

0.18

0.17 0.16

0.16

0.15

0.14

0.13

0.13

0.13

0.12

0.12

0.11

0.11

0.1

0.08

0.07

0.00

0.25

0.50

0.75

1.00

Sensi

tivity

0.00 0.25 0.50 0.75 1.00

1 - Specificity

Points labeled by predicted probability

ROC Curve for WeightArea Under the Curve = 0.6367

0.00

0.25

0.50

0.75

1.00

Sensi

tivity

0.00 0.25 0.50 0.75 1.00

1 - Specificity

Weight (0.6367)Age (0.6985)

ROC Curve (Area)

ROC Curves for Comparisons


ROC Association Statistics

ROC Model

Mann-Whitney

Somers' D Gamma Tau-a Area Standard

Error

95% Wald

Confidence Limits

Age 0.6985 0.0504 0.5998 0.7972 0.3970 0.4065 0.0903

Weight 0.6367 0.0551 0.5287 0.7447 0.2734 0.2758 0.0622

ROC Contrast

Coefficients

ROC

Model Row1

Age 1

Weight -1

ROC Contrast Test Results

Contrast DF Chi-Square Pr > ChiSq

Reference = Weight 1 0.6822 0.4088

ROC Contrast Estimation and Testing Results by Row

Contrast Estimate

Standard

Error

95% Wald

Confidence Limits Chi-Square Pr > ChiSq

Age - Weight 0.0618 0.0748 -0.0848 0.2084 0.6822 0.4088

/* HW5, Problem 3 */

/*ods rtf file="SAS Output_HW5 Prob 3_files.rtf";*/

title1 'HW5 Problem 3';

data colds;

input colds total titer$ virus$ social$;

datalines;

25 33 f<=2f fRV39f f1-5f

20 38 f<=2f fRV39f f>=6f

18 30 f<=2f fHanksf f1-5f

21 43 f<=2f fHanksf f>=6f

11 34 f>=4f fRV39f f1-5f

8 42 f>=4f fRV39f f>=6f

3 26 f>=4f fHanksf f1-5f

3 30 f>=4f fHanksf f>=6f

;

run;

proc print;

run;

proc logistic data=colds outest=betas covout;

title2 'Stepwise Regression on Cold Data';

class titer virus social/param=ref;

model colds/total = titer virus social titer*virus virus*social titer*social

/ selection=stepwise

slentry=0.05

slstay=0.05

details

lackfit;

output out=pred p=phat lower=lcl upper=ucl stdreschi = q reschi=p h=h xbeta=eta predprob=(individual crossvalidate);

run;

/* Creating the standardized residuals from the "pred" dataset */

data pred2; set pred; res = p/sqrt(1-h); run;

/* Usual plot of stand. Residuals vs predicted eta_i */

proc sgplot data=pred2;

title2 "Residuals vs predicted eta_i";

scatter y=res x=eta;

run;

/* Now, doing the LOESS overlay on the r_i vs eta_i fit. You can learn about smoothing parameter selection via the link below

http://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_loess_sect004.htm

Lot to learn! */

proc loess data=pred2;

title2 "Residuals vs predicted eta_i with LOESS Overlay";

model res = eta;

run;

/* Also, you can try this one below */

proc sgscatter data=pred;

title2 "Std. Pearson residual plots";

plot p*eta q*eta / loess;

run;

/*ods rtf close;*/

/* These are some extras you can ignore */

proc print data=betas;

title2 'Parameter Estimates and Covariance Matrix';

run;

proc print data=pred;

title2 'Predicted Probabilities and 95% Confidence Limits';

run;

/* HW5, problem 6 */

/*ods rtf bodytitle file='SAS Output HW5 Problem 6.rtf';*/

title 'HW5, problem 6';

data vaso;

input cons volume rate;

datalines;

/* DATALINES not shown here. */

.

.

.

;

run;

data vaso1; set vaso;

lvol = log(volume);

lrate = log(rate);

run;

/*proc print data=vaso1;*/

/*run;*/

proc logistic data=vaso1 descending;

title2 'Vaso Data Analysis [Logit]';

model cons = lvol lrate/link = logit aggregate lackfit;

output out=pred p=phat lower=lcl upper=ucl reschi=p h=h xbeta=eta;

run;


title2 'Vaso Data Analysis [Probit]';

model cons = lvol lrate/link = probit aggregate lackfit;


run;


title2 'Vaso Data Analysis [Cloglog]';

model cons = lvol lrate/link = cloglog aggregate lackfit influence iplots;


run;

/* Note that the 2 abberrant observations are 4 ans 18. Both have high Pearson, Dbbeta, and Cis. */

data vaso2; set vaso1;

if _n_=4 then delete; if _n_=18 then delete;

proc logistic descending data=vaso2;

model cons = lvol lrate / link=cloglog lackfit;

run;

ods rtf close;

/* Aside */


model cons = lvol lrate/link = cloglog aggregate lackfit influence;

output out=pred reschi = u stdreschi=r xbeta=eta p=p;

run;

data pred1; set pred; keep r u;

run;

proc print data=pred1;

run;

title;

/* Problem 7, Agresti 6.4 */

/*a = alcohol; c = cigarette use; m = marijuana use; r = race [1 = white, 2 = black]; g = gender [1 = Female, 2 = male]; count = counts (binomial);*/

title "Problem 7, Agresti 6.4";


data mari;

input a c m r g count total;

datalines;

1 1 1 1 1 405 673

1 1 1 2 1 23 46

1 2 1 1 1 13 231

1 2 1 2 1 2 21

2 1 1 1 1 1 18

2 1 1 2 1 0 1

2 2 1 1 1 1 118

2 2 1 2 1 0 12

1 1 1 1 2 453 681

1 1 1 2 2 30 49

1 2 1 1 2 28 229

1 2 1 2 2 1 19

2 1 1 1 2 1 18

2 1 1 2 2 1 9

2 2 1 1 2 1 134

2 2 1 2 2 0 17

;

run;

proc logistic data= mari outest=betas covout;

title2 'Stepwise Regression on Marijuana Data';

class a c r g/param=ref;

model count/total = a|c|r|g@2 / selection=stepwise slentry=0.05 slstay=0.05 details lackfit;

output out=pred p=phat lower=lcl upper=ucl stdreschi = q reschi=p h=h xbeta=eta predprob=(individual crossvalidate);

run;

/* Creating the standardized residuals from the "pred" dataset */

data pred2; set pred; res = p/sqrt(1-h);

/* Usual plot of stand. Residuals vs predicted eta_i */

proc sgplot data=pred2;

title2 "Residuals vs predicted eta_i";

scatter y=res x=eta;

run;

proc loess data=pred2;

title2 'residuals vs. eta';

model res = eta;

run;

/* Also, you can try this one below */

proc sgscatter data=pred;

title2 "Std. Pearson residual plots";

plot p*eta q*eta / loess;

run;

/*ods rtf close;*/

/* Problem 8, heart.sas */

title "Problem 8, heart.sas";


data heart;

input ID AG S D Ch H W CNT garbage;

datalines;

/* DATALINES not shown here (very long) */

.

.

.

;

run;

/* (a) Setpwise elimination, you can also try Backward Elimination */

proc logistic data=heart descending;

title2 'Stepwise Regression on Heart Data';

model CNT = AG|S|D|Ch|H|W@2

/ selection=stepwise

slentry=0.05

slstay=0.05

details

lackfit

scale = none;

output out=predict p=phat lower=lcl upper=ucl stdreschi = q reschi=p h=h2 xbeta=eta c = c predprob=(individual crossvalidate);

run;

proc logistic data=heart descending;

title2 'Backward Regression on Heart Data';

model CNT = AG|S|D|Ch|H|W@2

/ selection=B fast ctable

slentry=0.05

slstay=0.05

details

lackfit;

output out=predict p=phat lower=lcl upper=ucl stdreschi = q reschi=p h=h2 c = c xbeta=eta predprob=(individual crossvalidate);

run;

/* Plots, and also comment from the Table on Outliers. Interpretation in details, wrt odds ratios */

proc sgscatter data=predict;

title2 "Cook's distance and Std. Pearson resids";

plot q*eta q*id c*id/loess ;

proc print data=predict(where=(c>0.2 or q>3 or q<-3));

*var y width color c r;

run;

/* Now, predictive accuracy */

proc logistic data = heart descending plots(only)=(roc(id=obs) effect);

model CNT = AG W/scale = none details lackfit rsquare;

run;

proc logistic data = heart descending plots;

model CNT = AG W/scale = none details lackfit rsquare outroc=ROCcurve ctable;

run;

/* Below find individual ROC curves for each covariate */

proc logistic data=heart plots=roc(id=prob);

model CNT(event='1') = AG W / nofit;

roc 'Age' AG;

roc 'Weight' W;

roccontrast reference('Weight') / estimate e;

run;

/*ods rtf close;*/

/* The GAM below is some extras */

*proc gam plots(clm);

proc gam data=heart;

model CNT(EVENT='1') = spline(AG) spline(Ch) spline(W) / dist=binomial link=logit;

run;

*ods graphics off;

*ods html close;

BIOS 625 HW 5 Solutions Sheet - dbandyop/BIOS625/HW/hw52015_Soln.pdf · BIOS 625 HW #5 Solutions...

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Transcript of BIOS 625 HW 5 Solutions Sheet - dbandyop/BIOS625/HW/hw52015_Soln.pdf · BIOS 625 HW #5 Solutions...